91Ó°ÊÓ

The product of the reactant concentration divided by the product concentration yields the dissociation constant.

Product of concentration of reactants\( \div \)concentration of products

Consider the reaction,

Since, it is given that,

Initial concentration of\({\rm{Zn}}({\rm{OH}})_4^{2 - }{\rm{\;is\;}}0.0465{\rm{M}}\)and dissociation constant\({K_{\rm{d}}} = 3.4 \times {10^{ - 15}}\),

Calculate the equilibrium concentration of\({\rm{Z}}{{\rm{n}}^{2 + }}{\rm{\;and\;O}}{{\rm{H}}^ - }\),

Use the dissociation constant formula,

\({K_d} = \frac{{\left( {Z{n^{2 + }}} \right) \cdot {{\left( {O{H^ - }} \right)}^4}}}{{\left( {Zn(OH)_4^{2 - }} \right)}}\)

\(3.4 \cdot {10^{ - 15}} = \frac{{x \cdot {{(4x)}^4}}}{{0.0465 - x}}\)

Since,\({K_d}\)is smaller than\({10^{ - 4}}\), assume that\(0.0465 - {\rm{x}} \approx 0.0465\)

\(3.4 \cdot {10^{ - 15}} = \frac{{x \cdot {{(4x)}^4}}}{{0.0465}}\)

\(256{x^5} = 1.58 \cdot {10^{ - 16}}\)

\({x^5} = 6.18 \cdot {10^{ - 19}}\)

\(x = 2.28 \cdot {10^{ - 4}}{\rm{M}}\)

Thus,

\(\begin{array}{*{20}{c}}{\left( {Z{n^{2 + }}} \right) = x = 2.28 \cdot {{10}^{ - 4}}{\rm{M}}}\\{\left( {O{H^ - }} \right) = 4x = 9.12 \cdot {{10}^{ - 4}}{\rm{M}}}\end{array}\)

Therefore, the equilibrium concentrations of \({\rm{Z}}{{\rm{n}}^{2 + }}{\rm{and\;O}}{{\rm{H}}^ - }\)are,

\(\begin{array}{*{20}{c}}{\left( {Z{n^{2 + }}} \right) = 2.28 \cdot {{10}^{ - 4}}{\rm{M}}}\\{\left( {O{H^ - }} \right) = 9.12 \cdot {{10}^{ - 4}}{\rm{M}}}\end{array}\)